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cestrela7 [59]
3 years ago
15

The compound cyanogen (CN)2 has covalent bonds shown in this diagram. Starting at left, an N connected by three lines, one above

the other, to a C connected by one line to a C connected by three lines, one above the other, to an N. How many electrons are shared between one nitrogen atom and one carbon atom? two three six seven
Physics
2 answers:
Serga [27]3 years ago
8 0

Answer:

C  For a 100% fact!!!!   :)

labwork [276]3 years ago
4 0

Answer:

I did C on edge and got this correct sooo

Explanation:

You might be interested in
edward travels 150 kilometers due west and then 200 kilometers in a direction 60 degrees north of west.what is his displacement
Naily [24]

The displacement of Edward in the westerly direction is determined as 338.32 km.

<h3>What is displacement of Edward?</h3>

The displacement of Edward can be determined from different methods of vector addition. The method applied here is triangular method.

The angle between the 200 km north west and 150 km west = 60 + 90 = 150⁰

The displacement is the side of the triangle facing 150⁰ = R

R² = a² + b² - 2abcosR

R² = 150² + 200²  - (2x 150 x 200)xcos(150)

R² = 62,500 - (-51,961.52)

R² = 114,461.52

R = 338.32 km

Learn more about displacement here: brainly.com/question/321442

#SPJ1

3 0
2 years ago
A grape is thrown straight up in the air. If it was thrown at 2 m/s, how high will it go in the air?
ser-zykov [4K]

Answer:

Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>

Explanation:

At the highest point, it is called the maximum height.

• From third newton's equation of motion:

{ \rm{ {v}^{2}  =  {u}^{2}  + 2gs}}

• At maximum height, v is zero

• u is initial speed

• g is -9.8 m/s²

• s is the height

{ \rm{0 {}^{2}  =  {2}^{2}   -  (2 \times 9.8 \times s)}} \\  \\ { \rm{4 = 19.6s}} \\  \\ { \rm{s = 0.204 \: m}}

4 0
3 years ago
Read 2 more answers
PLZ Help Me I give you brainlist also NO LONKS OR I REPORT and Have a gr8 day my peeps
Sergio039 [100]

Answer:

the answer is a because I saw it in a syllabus

3 0
3 years ago
An example for curvilinear motion.
Shkiper50 [21]

Ball thrown into the air at an angle.

5 0
3 years ago
What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec
jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

E = \sigma T^{4}  (1)

Where \sigma is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

\lambda max = \frac{2.898x10^{-3} m. K}{T}   (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}  (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

T = \frac{2.898x10^{-3} m. K}{\lambda max}   (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), \lambda max (450 nm) will be express in meters:

450 nm . \frac{1m}{1x10^{9} nm}  ⇒ 4.5x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}

T = 6440 K

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

700 nm . \frac{1m}{1x10^{9} nm}  ⇒ 7x10^{-7}m

T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}

T = 4140 K

Comparison:

\frac{6440 K}{4140 K} = 1.55

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0
4 years ago
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